Chemistry: Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. determine the percentage of unit cell volume that is occupied in (a) a face- centered cubic lattice, (b) a body-centered cubic lattice, and (c) a diamond lattice.

Assume that each atom is a sphere, and that the - id4784570, murdocksfamilyouoxvm

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jngonzo1226

16.01.2022

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05% The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03% The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. determine the percentage of unit cell volume that is occupied in (a) a face- centered cubic lattice, (b) a body-centered cubic lattice, and (c) a diamond lattice.

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